In the right-angled triangles ABC and A1B1C1 from the vertices of the right angles C and C1

In the right-angled triangles ABC and A1B1C1 from the vertices of the right angles C and C1, the heights CH and C1H1 are drawn, CH = C1H1, AH = A1H1. Prove that triangles ABC and A1B1C1 are equal.

Consider triangles ABН and A1B1H1. By hypothesis, CH = C1H1, AH = A1H1, then, based on the Pythagorean theorem, AB = A1B1.

According to the properties of the height of a right-angled triangle, AH ^ 2 = BH * CH, A1H12 = B1H1 * C1H1.

Since by the condition CH = C1H1, and AH = A1H1, then BH = B1H1, then BC = B1C1.

Then triangles ABC and A1B1C1 are equal in legs AB and A1B1 and hypotenuses BC and B1C1.